# where did I go wrong (phasor calculations)?

Discussion in 'Homework Help' started by Berticus, Apr 11, 2009.

1. ### Berticus Thread Starter New Member

Dec 12, 2008
2
0
Attached is the schematic.

$Z_C = -j2000 \\
Z_L = j1000$

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$56.56 = (1500 + j1000)i_1 - j1000i_2 \\
0 = -j1000i_1 + (1000 - j1000)i_2$

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$1000\angle90^{\circ}i_1 = 1414.21\angle-45^{\circ}i_2 \\
1_1 = 1.41421\angle-135^{\circ}i_2$

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$56.56 = 1802.78\angle33.69^{\circ}(1.41421\angle-135^{\circ})i_2 - j1000i_2 \\
= 2549.51\angle-101.31^{\circ}i_2 - j1000i_2 \\
= (-500 - j2500)i_2 - j1000i_2 \\
= (-500 - j3500)i_2 \\
= 3535.53\angle261.87^{\circ}i_2 \\
i_2 = 0.01600\angle98.1301^{\circ} = -0.0022624 + j0.01357$

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$i_1 = 1.41421\angle-135^{\circ}(0.01600\angle98.1301^{\circ}) \\
= 0.022624\angle-36.8699^{\circ} = 0.018099 - j0.01357$

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$I_s = i_1 = 0.022624\angle-36.8699^{\circ} \\
I_C = i_2 = 0.01600\angle98.1301^{\circ} \\
I_L = i_1 - i_2 = 0.018099 - j0.01357 + 0.0022624 - j0.015837 \\
= 0.0203614 - j0.029407 \\
= 0.03577\angle55.301^{\circ}$

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$I_C$ is the only one that is correct. The other two are out of phase by $180^\circ$. However, their magnitudes are correct.

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Last edited: Apr 11, 2009
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
11-12th line down you have

56.56 = (3535.53 @ 261.87°) i2

hence

i2= 0.016 @ 98.13°

why not .....

i2 = 0.016 @ -261.87°

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
No problem - your values are OK for i2 since +98.13° is the same as -261.87°.

I think the problem occurs when you convert the polar form of i2 to rectangular form.

I have ....

i2 = .016 @ angle(98.13°) = -.00226 +j 0.01584

whereas you have ....

i2 = -.00226 +j 0.01357